"The Ghost In The Machine" <ewill3@earthlink.net> wrote in message
news:pan.2006.03.20.04.54.57.5072@earthlink.net...

On Sun, 19 Mar 2006 17:21:17 -0500, Spaceman wrote:

"The Ghost In The Machine" <ewill3@earthlink.net> wrote in message
news:pan.2006.03.19.19.14.22.591012@earthlink.net...

There is some dispute as to whether lightspeed is constant among the
non-scientists. (Among scientists the issue is pretty much settled, and
has been since 1983.)

So, Mr scientist,
Please do show why a speed of c would be c still for an observer heading
towards the speeding wavefront? Lets say the observer is traveling at
0.5c
towards the source where the lightwave left.

I'm more a mathematician than a scientist, but I can refer you to

http://www.fourmilab.ch/etexts/einstein/specrel/www/

where a translated version of Einstein's paper is available. In it, he
makes the then-revolutionary hypothesis that lightspeed is c everywhere,
and derives the Lorentz.

The reverse is easier: given the Lorentz, derive constancy of lightspeed.
In short, assume O has a light source, and A is moving at v = 0.5c
relative to that source, and that the coordinate systems of O and A are
related by the Lorentz:

t_A = (t_O - v * x_O/c^2) / sqrt(1-v^2/c^2)
x_A = (x_O - v * t_O) / sqrt(1-v^2/c^2)
y_A = y_O
z_A = z_O

This Lorentz is invertible, yielding:

t_O = (t_A + v * x_A/c^2) / sqrt(1-v^2/c^2)
x_O = (x_A + v * t_A) / sqrt(1-v^2/c^2)
y_O = y_A
z_O = z_A

(the proof is left to the interested reader but isn't horribly difficult;
substitute one into the other and derive an identity).

At time t_O = 0 we have O emit a light pulse from his origin.
In O's coordinate space this light pulse travels at velocity c in
a spherical wavefront:

x_O^2 + y_O^2 + z_O^2 - c^2*t_O^2 = 0

How does A see this light pulse? By performing the obvious substitutions,
one gets the equation

g^2*(x_A+v*t_A)^2+y_A^2+z_A^2-c^2*g^2*(t_A+v*x_A/c^2)^2=0

where I've written g = 1/sqrt(1-v^2/c^2) for brevity,
and then

g^2*(x_A^2+v^2*t_A^2+2*v*x_A*t_A)+y_A^2+z_A^2
-c^2*g^2*(t_A^2*v^2*x_A^2/c^4-K^2/g^2+2*t_A*v*x_A/c^2)=0

g^2*(x_A^2+v^2*t_A^2)+y_A^2+z_A^2
-c^2*g^2*(t_A^2+v^2*x_A^2/c^4)=0

g^2*x_A^2-c^2*g^2*v^2*x_A^2/c^4+y_A^2+z_A^2+g^2*v^2*t_A^2-c^2*g^2*t_A^2=0

g^2*x_A^2*(1-v^2/c^2)+y_A^2+z_A^2-c^2*g^2*t_A^2*(1-v^2/c^2) = 0

Since g^2*(1-v^2/c^2) = 1, we get

x_A^2+y_A^2+z_A^2-c^2*t_A^2 = 0

and therefore A also sees the light pulse as receding from him at speed c
from *his* origin.

I could set up a light pulse in O's space not at his origin and not at
time 0, if you like, but the math gets much hairier. However, if O sets
up a pulse of light with the equation

(x_O-A_O)^2 + (y_O-B_O)^2 + (z_O-C_O)^2 - c^2*(t_O-D_O)^2 = 0

then at the end of the day (and a *lot* of grinding) one gets

(x_A-A_A)^2 + (y_A-B_A)^2 + (z_A-C_A)^2 - c^2*(t_A-D_A)^2 = 0

where

D_A = (D_O - v * A_O/c^2) / sqrt(1-v^2/c^2)
A_A = (A_O - v * D_O) / sqrt(1-v^2/c^2)
B_A = B_O
C_A = C_O

Make of it what one will, but the math checks out at least, if one is
willing to grind it out.

Ghost,
The math would also check out with 50 meters per second is a constant.

It is silly trick math that is self limiting at best and the only thing it
is good for is tricking people into thinking lightspeed is constant to all.
Change the c to 50 meters per second and then all of sudden
50 meters per second is constant to all also.
I would think someone with such a mathematical background
you seem to show could figure out why any "transform" is simply
a math trick to limit the outcome.

The problem is that silly trick math would not account
for why a 1 second light pulse from the source
would be a shorter duration for the observer heading
towards the source at 0.5c and yet silly basic algebra
would.

So.
If a light source turned on, then off for a total
of a 1 second light pulse and was heading away from the source.
and an observer was heading towards the source,
how long (time) would the pulse be lit for such observer?
Do you think you need all that math to actually find out
how long the duration of the lightpulse would be for the observer?
Hint: you don't.
You simply need basic algebra only with no transform at all.
It works on all relative speeds.

As for the scientists -- there are a fair number of experiments done
regarding various parts of relativity.

Yup,
and for the constant to all lightspeed bologna,
I have gone through any I could find data for, and not one of them
physically measured the relative light speed for an observer
heading towards the source.

http://en.wikipedia.org/wiki/Special_relativity

might be helpful, although I see more links to textbooks than actual
experiments.

Been there,
Done that.
Lightspeed is a speed and there is no physical way a speed
can be constant to all without resorting to mathematical tricks
such as a transform.

Again
Try it.
c = 50 meters per second.
and see what happens to any transform using that speed as c.
:)

--
James M Driscoll Jr
Spaceman